Alternate Definition of a Group

A set G with binary composition denoted multiplicatively is a group if

(i) The composition is associative.
(ii) For every pair of elements a,b \in G, the equations ax = b and ya = b have unique solutions in G.

Proof: Binary operation implies that the set G, under consideration is closed under the operation. Now to prove that the set G is a group we have to show the left identity exists and each element of G possesses left inverse with respect to the operation under consideration.
It is given that for every pair of elements a,b \in G the equation ya = b has a solution in G. Therefore, say e \in G such that e \cdot a = a.
Now, let us suppose that b is any arbitrary element of G. Therefore, there exist e \in G such that ax = b.

b =  ax
  \Rightarrow eb = e\left( {ax} \right)
  \Rightarrow eb = \left( {ea} \right)x
\,  \Rightarrow eb = ax
  \Rightarrow eb = b

Therefore there exist e  \in G such that eb =  b, \forall b \in G
\therefore e is the left identity.

Now, let b = e \in G be an element. ya = b \Rightarrow ya = e  \Rightarrow y is inverse of a in G.
Let y = {a^{ - 1}} such that {a^{ - 1}}a = e, then {a^{ - 1}} \in G as ya = e has got solution in G.

Thus {a^{ - 1}} is the left inverse of a in G. Therefore each element of possesses left inverse. Hence G is a group for the given composition if the postulate and (ii) are satisfied.